Object a has a mass of 20 g

Planet with an atmosphere that rains sulfuric acid

Paladinen [302]

Answer:

Venus

Explanation:

Venus is the second plate in the solar system. It is a terrestrial planet and it is part of the inner rocky planets.

In Venus, it rains sulfuric acid but the rain never reaches the surface before it becomes evaporated. The acid forms from the combination of sulfur oxide and water in the atmosphere at a height of about 42km. As it condenses and falls, it becomes evaporated back at lower elevations. The surface is therefore protected from the sulfuric acid rain.

The sulfur oxide and water vapor must have been derived from volcanic activities in geologic times past.

3 0

3 years ago

A car is travelling at 15 m/s on a horizontal road and stopped after 4 s. The coefficient of kinetic friction between the tires

givi [52]

Answer:

Fr,= umg

umg= ma

a= v/t

umg= mv/t

u= v/gt= 0.38

3 0

2 years ago

A 500 kg roller coaster moving at 15 m/s suddenly comes to a stop at the end of the ride. How much work energy was needed to sto

saveliy_v [14]

Answer:

energy required=-energy lost

energy lost=change in kinetic energy

EL=1/2 mv^2

4 0

3 years ago

When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f

Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer ( $r$ ), measured in kilometers, can be represented by a right triangle:

$r = \sqrt{x^{2}+y^{2}}$ (1)

Where:

$x$ - Horizontal distance between the rocket and the observer, measured in kilometers.

$y$ - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket ( $\theta$ ), measured in sexagesimal degrees, is defined by the following trigonometric relation:

$\tan \theta = \frac{y}{x}$ (2)

If we know that $x = 5\,km$ , then the expression is:

$\tan \theta = \frac{y}{5}$

And the rate of change of this angle is determined by derivatives:

$\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y$

$\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}$

$\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}$

$\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}$

Where:

$\dot \theta$ - Rate of change of the angle of elevation, measured in sexagesimal degrees.

$\dot y$ - Vertical speed of the rocket, measured in kilometers per hour.

If we know that $y = 4\,km$ and $\dot y = 400\,\frac{km}{h}$ , then the rate of change of the angle of elevation is:

$\dot \theta = 48.780\,\frac{\circ}{s}$

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0

3 years ago

Name the quantity which is measured by the area occupied below the velocity time graph

kiruha [24]

If you mark off a beginning time and ending time on the graph,
then the area under the part of the graph between those limits
is the distance covered during that period of time.

3 0

3 years ago

Read 2 more answers